How number nifty are you?

[KLin]

Originally posted by: FoBoT
i read the first question and closed my browser

Same here .

 

[thesurge]

Originally posted by: DrPizza

Originally posted by: TuxDave
Edit: Ok I know where I screwed up. "My idea is that your three digit number can never contain a zero because you would end up with non-unique digits. "

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

I concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Where the different type is the average of the other two. (eg OOE, the even number is the average of the two odd numbers)

Looking at the combination of OO I get 20. 1,3,5,7,9 for the first number and the second number can only have 4 choices.

Then you multiply 3 since OOE,OEO,EOO have the same number of choices to get 60.

EE (where the first digit isn't in the hundred's place) you get 0,2,4,6,8 so 5x4=20
EE (where the first digit is in the hundred's place) you get 4*4=16
So OEE,EOE,EEO you get 20+16+16=52

52+60=112. Yay.

<-- nerd

I think you looked at it the hard way...
I started with 0, and didn't look for what you concluded, although it sort of appears.
0&1? nope
0&2? Yes, avg = 1
0 & 3? Nope.

At this point, I started counting: distinct two digit pairs such that the 3rd digit will be their average.
02, 04, 06, 08, 24, 26, 28, 46, 48, 68
That takes care of the even ones;
And, for the odds,
13, 15, 17, 19, 35, 37, 39, 57, 59, 79

(note: I quickly realized there would be 4+3+2+1 combinations of evens and the same number of combinations of two odds, hence I was at 20 without actually thinking out all of the pairs, or writing them all down)

With the third digit being the average of the two existing digits, it created 20 different three digit sets. There are 6 different orders for almost each of these (in a moment)
i.e. for 39, the third digit is 6. You can have 369, 396, 639, 693, 936, 963

So, 20*6 = 120.
But, then, 024 really isn't a three digit number. 24 is a two digit number. For each of the pairs that contain a 0, there are two ways to arrange them so that the 0 is first. That's 4*2=8. So, my final total is 112 also.

btw. Since I agree with Tuxdave on the answer of 112, the test is incorrect.

If you look at my solution, I also got 112. And if you look closely at the Forbes page for the problem, they also got 112. :Q

 

[Mrfrog840]

thank you for making me feel even stupider

 

[chuckywang]

LOL, this is so easy. If I had paper, I could ace it.

 

[KingGheedora]

Originally posted by: TuxDave

I concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Shouldn't it be either EEE, or EOO, OEO, OOE?

Because the third # (the # that is the avg of the other two #'s) is the result of division by 2, we know the third number is always even.

The first two numbers, whose avg is the third, have to add up to an even number (in order for them to be divisible by two, to calculate the average). You can only get an even number as the result of addition of two numbers if you have either E + E, or O + O.

Given those two truths, the numbers that satisfy the problem have to be in one of these forms: EEE, EOO, OEO, OOE.

 

[TheRyuu]

Yea, I remember taking this not to long ago.

It was pretty hard.

(took the AMC12 AFAIK)

 

[thesurge]

Originally posted by: KingGheedora

Originally posted by: TuxDave

I concluded that you can only end up with these forms.

O = Odd
E = Even

OOE
OEO
EOO
EEO
EOE
OEE

Shouldn't it be either EEE, or EOO, OEO, OOE?

Because the third # (the # that is the avg of the other two #'s) is the result of division by 2, we know the third number is always even.

The first two numbers, whose avg is the third, have to add up to an even number (in order for them to be divisible by two, to calculate the average). You can only get an even number as the result of addition of two numbers if you have either E + E, or O + O.

Given those two truths, the numbers that satisfy the problem have to be in one of these forms: EEE, EOO, OEO, OOE.

Yes, the two numbers must have the same parity, but just because they sum to an even number does not mean the average must be even (they may have only one 2 in the prime factorization).
Ex. 456 (EOE)
Also, TuxDave is incorrect. We can have form:
Ex. 579 (OOO)
but he ends up with the write answer (maybe coincidently).

 

[DrPizza]

<- merely acknowledging that in addition to tuxdave, thesurge was also correct. (And so was the answer on the exam; it took forever to go through that exam on dial-up. Not too bad though.)

 

[MidasKnight]

Originally posted by: 40Hands
I suck at math. Real bad.

:thumbsup:

 

[eplebnista]

Originally posted by: KLin

Originally posted by: FoBoT
i read the first question and closed my browser

Same here .

thirded

 

[DrPizza]

Originally posted by: MidasKnight

Originally posted by: 40Hands
I suck at math. Real bad.

:thumbsup:

Wow, two fools who are proud of being mathematically illiterate.